91Ó°ÊÓ

To identify the convergence properties of the given function, let's first switch to the iterated integral form: \[\int_{a}^{\frac{1}{a}} du \int_{0^+}^{1} (\ln ut) \, dt.\]

We can apply Fubini's theorem since the function \(\ln ut\) is continuous, to switch the order of integration: \[\int_{0^+}^{1} dt \int_{a}^{\frac{1}{a}} (\ln ut) \, du = \int_{0^+}^{1} t \ln t \int_{a}^{\frac{1}{a}} \frac{du}{t}.\] Now, if we let \(v = \ln t\), the integral becomes: \[\int_{-\infty}^{0} v e^v dv.\]

Using integration by parts, with \(u = v\) and \(dv = e^v dv\), we obtain: \[\int_{-\infty}^{0} v e^v dv = \left. -ve^v \right|_{-\infty}^{0} - \int_{-\infty}^{0} -e^v dv = -\left. e^v \right|_{-\infty}^{0} = 1.\] So, \(\int_{0^+}^{1} (\ln t) \, dt = 1\). (ii) For the improper integrals \(\int_{0^+}^{1} e^{-ut}(\sin t) t^{-p} \, dt\) and \(\int_{0^+}^{1} e^{-ut}(\cos t) t^{-p} \, dt\), we want to show that they converge uniformly for \(u \in [0, \infty)\). The approach for these parts will be similar to (i). This time, consider the condition on the exponent of the \(t\)-term and the uniformity condition. (iii) For improper integrals \(\int_{0^+}^{1} \sin(u / t) t^{-q} \, dt\) and \(\int_{0^+}^{1} \cos(u / t) t^{-q} \, dt\), we want to show that they converge uniformly for \(u \in (-\infty, \delta] \cup [\delta, \infty)\). Similar to the previous part, focus on the exponentiation of the \(t\)-term and the uniformity condition. Apply the necessary convergence tests and discuss the final result. (iv) Finally, for improper integrals \(\int_{0^+}^{1} e^{-u / t} \sin(1 / t) t^{-q} \, dt\) and \(\int_{0^+}^{1} e^{-u / t} \cos(1 / t) t^{-q} \, dt\), we want to show that they converge uniformly for \(u \in [\delta, \infty)\). Similar to (iii), focus on the exponentiation of the \(t\)-term and the uniformity condition. Use the given hints and the knowledge from the previous parts to prove the convergence.